Why this equation cannot be equal to 1?

Question

How can this be solved ?

$$ 2e^{\frac{x}{3}} - e^{-\frac{x}{2}} = 1$$

Answer 1

Let $f(x) = 2e^{\frac{x}{3}} - e^{-\frac{x}{2}}$.

Then $f'(x) = \frac{2}{3}e^{\frac{x}{3}} + \frac{1}{2}e^{-\frac{x}{2}} > 0 $ for all $x \in \mathbb{R}$ so $f(x)$ is strictly increasing.

Therefore, there is at most one solution to $f(x) =1$ and I see one, $x = 0$

Answer 2

Change the variable: define

$$X:=e^{\frac x 6}.$$

Then you can rewrite your equation:

$$x=0\text{ or }2X^2-\frac 1{X^3}=1.$$

The second part is equivalent to

$$2X^5-1=X^3$$

iff

$$2X^5-X^3-1=0.$$

You get one obvious solution $X=1$, and you can now factorize it by $(X-1)$ and you get:

$$(X-1)(2X^4+2X^3+X^2+X+1)=0$$

and $2X^4+2X^3+X^2+X+1$ has not real solution.

And $X=1$ gives you what we have already find: $x=0$.

Finally, there is only one solution: $x=0$.